Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\tan x - 1}}{{x - \pi /4}} \cr
& = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\frac{{\sin x}}{{\cos x}} - 1}}{{x - \pi /4}} \cr
& = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\cos x\left( {x - \pi /4} \right)}} \cr
& \cr
& {\text{Let }}t = x - \pi /4 \cr
& x \to \pi /4 \Rightarrow t = \pi - \pi = 0 \cr
& t \to 0{\text{ when }}x \to \pi /4 \cr
& \cr
& {\text{Substituting}} \cr
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\cos x\left( {x - \pi /4} \right)}} = \mathop {\lim }\limits_{t \to 0} \frac{{\sin \left( {t + \pi /4} \right) - \cos \left( {t + \pi /4} \right)}}{{\cos \left( {t + \pi /4} \right)t}} \cr
& = \mathop {\lim }\limits_{t \to 0} \frac{{\sin \left( {t + \pi /4} \right) - \cos \left( {t + \pi /4} \right)}}{{\cos \left( {t + \pi /4} \right)t}} \cr
& {\text{Where:}}\,\,\sin \left( {t + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\left( {\sin t + \cos t} \right){\text{ and }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {t + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\left( {\cos t - \sin t} \right) \cr
& \cr
& = \mathop {\lim }\limits_{t \to 0} \frac{{\frac{{\sqrt 2 }}{2}\left( {\sin t + \cos t} \right) - \frac{{\sqrt 2 }}{2}\left( {\cos t - \sin t} \right)}}{{\frac{{\sqrt 2 }}{2}\left( {\cos t - \sin t} \right)t}} \cr
& = \mathop {\lim }\limits_{t \to 0} \frac{{\sin t + \cos t - \cos t + \sin t}}{{\left( {\cos t - \sin t} \right)t}} \cr
& = \mathop {\lim }\limits_{t \to 0} \frac{{2\sin t}}{{\left( {\cos t - \sin t} \right)t}} \cr
& {\text{Use the product property of limits}} \cr
& = 2\left( {\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t}} \right)\left( {\mathop {\lim }\limits_{t \to 0} \frac{1}{{\cos t - \sin t}}} \right) \cr
& = 2\left( 1 \right)\left( {\frac{1}{{\cos 0 - \sin 0}}} \right) \cr
& = 2\left( 1 \right)\left( 1 \right) \cr
& = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\tan x - 1}}{{x - \pi /4}} = 2 \cr} $$
